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The volume of a spherical balloon is increasing at the rate of 25 cm3/sec. Find the rate of change of its surface area at the instant when the radius is 5 cm

Manvendra Singh chahar , 10 Years ago
Grade Upto college level
anser 2 Answers
SHAIK AASIF AHAMED

Last Activity: 10 Years ago

Hello student,
Let r,V and S be radius,volume and surface area of the sphere at time t.
v=(4/3) \pi r^{3} and S=4\pi r^{2}
Rate of change of volume =(dv/dt)=25
4\pi r^{2}dr/dt=25
Rate of change of surface area=(ds/dt)=8\pi r(dr/dt)=8\pi r(25/4\pi r^{2})=50/r
So rate of change of surface area when r is 5cm=(50/5)=10cm2/sec
Thanks and Regards
Shaik Aasif
askIITians faculty

Dev Desai

Last Activity: 5 Years ago

Volume of sphere = (4/3)(pi)(r)^3
Surface area of sphere =(4)(pi)(r)^2
 
V/S = ((4/3)(pi)(r)^3)/((4)(pi)(r)^2) =>  r/3
Therefore, ((dv/dt)×3)/r= (ds/dt)
Hence, ds/dt = (25×3)/5 = 10 cm^3/sec
 

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